Is Matrix Cells in Distance Order easy or hard?

Matrix Cells in Distance Order is classified as an Easy problem. The brute force solution requires simple matrix iteration and sorting, while the optimal bucket sort approach introduces a useful optimization based on the bounded range of Manhattan distances.

Matrix Cells in Distance Order Python/Java solution

In Python or Java, iterate through all grid coordinates, compute the Manhattan distance |r−r0| + |c−c0|, and either sort the list of cells by that value or place them into distance buckets. Python commonly uses list sorting with a key function, while Java uses a comparator or bucket lists.

How to solve Matrix Cells in Distance Order in O(n)?

Compute the Manhattan distance for every cell and store the coordinate in a bucket corresponding to that distance. Because the maximum distance is bounded by (rows−1)+(cols−1), you can iterate through the buckets sequentially to produce the result. This processes R*C cells once, giving O(R*C) time complexity.

What is the best approach for Matrix Cells in Distance Order?

The bucket sort method is the best approach. Since the Manhattan distance ranges from 0 to (rows−1)+(cols−1), you can group cells into distance buckets instead of sorting them. This reduces the runtime from O(R*C log(R*C)) to O(R*C) while still processing every cell exactly once.

Is Matrix Cells in Distance Order asked at Google/Amazon/Meta?

Matrix and grid traversal problems with Manhattan distance frequently appear in interviews at companies like Amazon and Google. While this exact problem may vary in wording, the underlying pattern of distance-based ordering in a grid is common in technical interviews.

What data structure is used in Matrix Cells in Distance Order?

The solution primarily uses arrays or lists to store coordinates. The optimal approach also uses an array of lists (buckets) where each index represents a Manhattan distance and stores all cells that share that distance.

What is the time complexity of Matrix Cells in Distance Order?

The optimal solution runs in O(R*C) time using bucket sort because each matrix cell is visited once and placed into a bucket indexed by its Manhattan distance. A simpler brute force solution that sorts all cells takes O(R*C log(R*C)) time.

Description Solution Video 22:02

1030. Matrix Cells in Distance Order

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You are given four integers row, cols, rCenter, and cCenter. There is a rows x cols matrix and you are on the cell with the coordinates (rCenter, cCenter).

Return the coordinates of all cells in the matrix, sorted by their distance from (rCenter, cCenter) from the smallest distance to the largest distance. You may return the answer in any order that satisfies this condition.

The distance between two cells (r₁, c₁) and (r₂, c₂) is |r₁ - r₂| + |c₁ - c₂|.

Example 1:

Input: rows = 1, cols = 2, rCenter = 0, cCenter = 0
Output: [[0,0],[0,1]]
Explanation: The distances from (0, 0) to other cells are: [0,1]

Example 2:

Input: rows = 2, cols = 2, rCenter = 0, cCenter = 1
Output: [[0,1],[0,0],[1,1],[1,0]]
Explanation: The distances from (0, 1) to other cells are: [0,1,1,2]
The answer [[0,1],[1,1],[0,0],[1,0]] would also be accepted as correct.

Example 3:

Input: rows = 2, cols = 3, rCenter = 1, cCenter = 2
Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]
Explanation: The distances from (1, 2) to other cells are: [0,1,1,2,2,3]
There are other answers that would also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]].

Constraints:

1 <= rows, cols <= 100
0 <= rCenter < rows
0 <= cCenter < cols

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1030. Matrix Cells in Distance Order

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